Respuesta :
Answer:
The answer is "Al= 9.8% and Fe=18.0%"
Explanation:
Given:
The weight of [tex]FeCl_3\ and \ ALCl_3[/tex] = 5.95g
[tex]gFeCl_3=gFe (\frac{Mw\ FeCL_3}{\text{atomic weight of Fe}})\\\\gAlCl_3=gAl (\frac{Mw\ AlCL_3}{\text{atomic weight of Al}})\\\\[/tex]
[tex]\to a = FeCl_3+AlCl_3\\\\\to a=x+y \\\\ \to a= 5.95[/tex]
[tex]\to a =x \ gFe (\frac{Mw\ FeCL_3}{\text{atomic weight of Fe}})+ y \ gAl (\frac{Mw\ AlCL_3}{\text{atomic weight of Al}})\\\\[/tex]
[tex]\to x (\frac{162.2}{55.85})+ y (\frac{133.34}{26.98})= 5.95\\\\\to 2.90x+4.94y=5.95\\\\\ similarly \ for \ oxidies:\\\\\to 143x+1.89y=2.62\\\\\to x= 1.07 \ \ \ and \ \ y= 0.58\\\\\to \ Al \% = \frac{0.58}{5.95} \times 100= \bold{9.8} \%\\\\\to \ Fe \% = \frac{1.07}{5.95} \times 100= \bold{18.0} \%[/tex]
The mass of the compound has been given as the mass of each element in the compound. The percent Fe in the mixture is 18%, and the percent Al in the mixture is 9.8%.
What is the percent composition?
The percent composition has been given as the mass of the element in the compound. The percent of iron and aluminum in their chlorides can be given as:
[tex]\rm Fe=Fe\;\times\;\dfrac{Mwt\;FeCl_3}{Mwt\;Fe}[/tex]
[tex]\rm Al=Al\;\times\;\dfrac{Mwt\;AlCl_3}{Mwt\;Al}[/tex]
The total mass of the compounds has been 5.95 grams.
Thus,
[tex]5.95=\rm FeCl_3+AlCl_3[/tex]
The mass of Fe and Al will be x and y of the mass in the compounds. Thus,
[tex]5.95=\rm x\;Fe\;\times\;\dfrac{Mwt\;FeCl_3}{Mwt\;Fe}\;+\;y\;Al\;\times\; \dfrac{Mwt\;AlCl_3}{Mwt\;Al}[/tex]
The oxides of the chemical has the sum of mass of 2.62 grams.
The mass of oxides can be given as:
[tex]2.62=\rm x\;Fe\;\times\;\dfrac{Mwt\;Fe2O_3}{Mwt\;Fe}\;+\;y\;Al\;\times\; \dfrac{Mwt\;Al_2O_3}{Mwt\;Al}[/tex]
Substituting the values of mass of Al, aluminum chloride, oxide, and Fe and its compounds.
[tex]\rm 2.90x+4.94y=5.95\\143x+1.89y=2.62[/tex]
Solving the equation, x = 1.07, and y = 0.58
The percent mass of the compounds has been given as the ratio of mass with compound mass.
Thus, the percent mass of Fe has been:
[tex]\rm Fe=\dfrac{1.07}{5.95}\;\times\;100\\ Fe=18\;\%[/tex]
The percent mass of Al has been:
[tex]\rm Al=\dfrac{0.58}{5.95}\;\times\;100\\ Al=9.8\;\%[/tex]
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